Chứng ninh bất phương trình: \(x^2-xy+y^2\ge0\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Thay x=-3 vào bpt, ta được:
\(\left(-3\right)^2-3\cdot\left(-3\right)+12=9+9+12>=0\)(luôn đúng)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(\dfrac{x-y}{-2}< x+y+1\)
\(\Leftrightarrow x-y>-2x-2y-2\)
\(\Leftrightarrow3x+y+2>0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: \(x\ne\dfrac{1\pm\sqrt{5}}{2}\)
TH1: \(x^2-x-1>0\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1+\sqrt{5}}{2}\\x< \dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)
\(\dfrac{\left|x^2-x\right|-2}{x^2-x-1}\ge0\)
\(\Leftrightarrow\left|x^2-x\right|-2\ge0\)
\(\Leftrightarrow\left|x^2-x\right|\ge2\)
\(\Leftrightarrow\left(\left|x^2-x\right|\right)^2\ge4\)
\(\Leftrightarrow x^4-2x^3+x^2-4\ge0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+2\right)\ge0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-1\end{matrix}\right.\)
TH2: \(x^2-x-1< 0\Leftrightarrow\dfrac{1-\sqrt{5}}{2}< x< \dfrac{1+\sqrt{5}}{2}\)
\(\dfrac{\left|x^2-x\right|-2}{x^2-x-1}\ge0\)
\(\Leftrightarrow\left|x^2-x\right|\le2\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+2\right)\le0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\le0\)
\(\Leftrightarrow-1\le x\le2\)
\(\Rightarrow\dfrac{1-\sqrt{5}}{2}< x< \dfrac{1+\sqrt{5}}{2}\)
Vậy \(S=[2;+\infty)\cup(-\infty;-1]\cup\left(\dfrac{1-\sqrt{5}}{2};\dfrac{1+\sqrt{5}}{2}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(\dfrac{x-2y}{2}>\dfrac{2x+y+1}{3}\)
\(\Leftrightarrow3x-6y>4x+2y+1\)
\(\Leftrightarrow x+8y+1< 0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\dfrac{-x^2-2x+1}{\left(x+2\right)\left(x-1\right)}>=0\)
=>\(\dfrac{x^2+2x-1}{\left(x+2\right)\left(x-1\right)}< =0\)
TH1: x^2+2x-1>=0 và (x+2)(x-1)<0
=>-2<x<1 và \(\left[{}\begin{matrix}x< =-1-\sqrt{2}\\x>=-1+\sqrt{2}\end{matrix}\right.\)
=>\(-1+\sqrt{2}< =x< 1\)
TH2: x^2+2x-1<=0 và (x+2)(x-1)>0
=>(x>1 hoặc x<-2) và \(-1-\sqrt{2}< =x< =-1+\sqrt{2}\)
=>\(-1-\sqrt{2}< =x< -2\)
\(x^2-xy+y^2=\left(x^2-2.x.\frac{y}{2}+\frac{y^2}{4}\right)+\frac{3y^2}{4}=\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}\ge0+0=0\forall x;y\)